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Tatiana [17]
3 years ago
5

Please help. I’ll mark you as brainliest if correct!

Mathematics
2 answers:
Delvig [45]3 years ago
6 0

Answer:

The two equations are

-2x-4y=20

-3x+5y=-25

multiply equation 1 by 5 and equation 2 by 4

-10x-20y=100

-12x+20y=-100

-22x=0

x=0

Substitute value in either equation

y=-5

So,option 1 is correct only one solution

Readme [11.4K]3 years ago
3 0

Answer:

\large \boxed{\sf \ \ x=0, \ \ y=-5 \ \ }

Step-by-step explanation:

Hello, please consider the following.

We have two equations:

(1) -2x - 4y = 20

(2) -3x + 5y = -25

5*(1)+4*(2) gives

   -10x - 20y -12x + 20y = 100 - 100 = 0

   -22x = 0

   x = 0

I replace in (1)

   -4y = 20

   y = -20/4 = -5

There is one solution x = 0, y = -5

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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A road-roller levels 3000m of road in 4hours How many metres does it level in a 24 hourday? please help me​
Naya [18.7K]

Answer:

18,000 meters

Step-by-step explanation:

We know:

  • v = 4 and t = 24

Substitute:

{ v = 4      ⇒ INTO FORMULA ⇒ D = V × T : D = 4 × 24

{ t = 24

Cross out the common factor:

⇒ 6 × 3000

Calculate the product or quotient:

⇒ 18,000

7 0
2 years ago
HELP!!!!!!!!!!!!!!!!!!!!!!ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Sphinxa [80]

Answer:72.70


Step-by-step explanation:  long story short math


6 0
3 years ago
Answer please I’m dying from math
charle [14.2K]

Answer:

\huge\boxed{\text{D)} \  15x^4 + 2x^3 - 8x^2 - 22x - 15}

Step-by-step explanation:

We can solve this multiplication of polynomials by understanding how to multiply these large terms.

To multiply two polynomials together, we must multiply each term by each term in the other polynomial. Each term should be multiplied by another one until it's multiplied by all of the terms in the other expression.

  • <em>We can do this by focusing on one term in the first polynomial and multiplying it by </em><em>all the terms</em><em> in the second polynomial. We'd then repeat this for the remaining terms in the second polynomial.</em>

Let's first start by multiplying the first term of the first polynomial, 3x^2, by all of the terms in the second polynomial. (5x^2+4x+5)

  • 3x^2 \cdot  5x^2 = 15x^4
  • 3x^2 \cdot 4x = 12x^3
  • 3x^2 \cdot 5 = 15x^2

Now, we can add up all these expressions to get the first part of our polynomial. Ordering by exponent, our expression is now

  • \displaystyle 15x^4 + 12x^3 + 15x^2

Now let's do the same with the second term (-2x) and the third term (-3).

  • -2x \cdot 5x^2 = -10x^3  
  • -2x \cdot 4x = -8x^2
  • -2x \cdot 5 = -10x
  • Adding on to our original expression: \displaystyle 15x^4 + 12x^3 - 10x^3 + 15x^2 - 8x^2 - 10x

  • -3 \cdot 5x^2 = -15x^2
  • -3 \cdot 4x = -12x
  • -3 \cdot 5 = -15
  • Adding on to our original expression: \displaystyle 15x^4 + 12x^3 - 10x^3 + 15x^2 - 8x^2 - 15x^2 - 10x - 12x - 15

Phew, that's one big polynomial! We can simplify it by combining like terms. We can combine terms that share the same exponent and combine them via their coefficients.

  • 12x^3 - 10x^3 = 2x^3
  • 15x^2 - 8x^2 - 15x^2 = -8x^2
  • -10x - 12x = -22x

This simplifies our expression down to 15x^4 + 2x^3 - 8x^2 - 22x - 15.

Hope this helped!

7 0
2 years ago
Read 2 more answers
4x-11y=68 ; 6x+5y=-27 by elimination
zimovet [89]

Answer:

x=\frac{1}{2}\\\\y=-6

Step-by-step explanation:

Given the following system of equations:

\left \{ {{4x-11y=68} \atop {6x+5y=-27}} \right.

In order to solve the system of equations using the Elimination Method, you can follow these steps:

- Multiply the first equation by -6 and the secondd equation by 4.

- Add both equations.

- Solve for "y".

Then:

\left \{ {{-24x+66y=-408} \atop {24x+20y=-108}} \right\\.........................\\86y=300\\\\y=\frac{-516}{86}\\\\y=-6

- Substitute the value of "y" into one of the original equations and solve for "x":

6x+5(-6)=-27\\\\6x=-27+30\\\\x=\frac{3}{6}\\\\x=\frac{1}{2}

4 0
3 years ago
Please help to solve this ​
Lerok [7]

Step-by-step explanation:

I think it will help you.

The image above will help you

7 0
2 years ago
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